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Does a surface such that any two different triangles on it are not congruent exist?

Added:Suppose that surfaces are defined by some continuous function:$R^2\rightarrow R$ and a triangle is a set of three non collinear points connected by three geodesics.

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That is, a metric space such that the map $(A,B,C)\mapsto (d(A,B),d(B,C),d(C,A))$ is injective? Then especially there cannot be an isosceles triangle. – Hagen von Eitzen Jan 6 '13 at 12:51
If your only limitation is metric spaces, then something like the powers of two on the real line would do the trick. Then even the function $(A, B) \mapsto d(A, B)$ is injective. – Arthur Jan 6 '13 at 13:25
Thank you! I had assumed a surface to be defined by some continuous function:$R^2\rightarrow R$. – user53216 Jan 6 '13 at 13:57
@user Please edit the question to make it precise: what is a surface, what is a triangle and what metric do you put on it. – user53153 Jan 6 '13 at 15:21
Try an triangle on a torus on the inside and on the outside – Willemien Oct 22 '14 at 7:43

Assuming by "surface" you mean a graph $\Gamma_f=\{(x_1,x_2,f(x_1,x_2))|\;x_1,x_2\in\mathbb R\}$ of a function $f:\mathbb R^2\to\mathbb R$, as you mention in the comments, and assuming by "different triangles on it are not congruent" you mean what Hagen von Eitzen says in the comments (i.e. that the function $\Phi$ we will define below is injective; edit: this in particular means that triangle should be interpreted as three point metric space, which might be a bit non-standard), the answer is negative.

The reason is as follows: we define the function $\Phi:\mathbb R^6\to\mathbb R^3$ (as suggested by Hagen von Eitzen in the comments) by $$\Phi(X,Y,Z)=(d(F(X),F(Y)),d(F(X),F(Z)),d(F(Y),F(Z))),$$ where $X,Y,Z\in\mathbb R^2$ and $F:\mathbb R^2\to \Gamma_f$ is the homeomorphism $(x,y)\to(x,y,f(x,y))$ (and we identify $\mathbb R^6$ with $\mathbb R^2\times\mathbb R^2\times\mathbb R^2$).

The problem is that $\Phi$ is a continuous function, but a continuous injection $\mathbb R^6\to\mathbb R^3$ cannot exist. This is because such an injection would in particular have to map $$S^5 =\{(x,y,z,u,v,w)\in\mathbb R^6|\;x^2+y^2+z^2+u^2+v^2+w^2=1\}$$ continuously and injectively into $\mathbb R^3$, and thus into $\mathbb R^5$. This would contradict the Borsuk-Ulam theorem.

Added: If by surface you mean $2$-manifold, the same argument works (and the answer is again negative): let $M$ be the $2$-manifold in question. (We assume that there is a metric given on $M$.) Then an arbitrary point in its interior has a neighborhood $U$ homeomorphic to $\mathbb R^2$. This homeomorphism $h:U\to\mathbb R^2$ allows us to define a metric $d_U$ on $\mathbb R^2$ by $d_U(h(p),h(q))=d(p,q)$, where $p,q\in U$ (and $d$ is the metric on $M$). Now define $\Phi:\mathbb R^6\to\mathbb R^3$ analogously to the definition above, i.e. $$\Phi(X,Y,Z)=(d_U(X,Y),d_U(X,Z),d_U(Y,Z)).$$ This is again continuous, and thus cannot be injective by Borsuk-Ulam.

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It seems like you are taking "congruence of triangles" to mean congruence of triangles in the ambient Euclidean space. I thought the question was asking about geodesic triangles. In the latter case there isn't an "SSS" rule for congruence, is there? – user29743 Jan 6 '13 at 20:12
@countinghaus: actually, I was interpreting "triangle" as a three point metric subspace. (Which I thought made sense, but I see it might be a non-standard use of the word.) The metric on $M$ is completely arbitrary, as long as it induces the correct topology. I agree that the question about geodesic triangles is a very interesting one. – Dejan Govc Jan 6 '13 at 20:23
Thank you!! : > – user53216 Jan 8 '13 at 12:12

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